# 0.1 Lecture 2: introduction to systems  (Page 7/8)

${v}_{o}\left[1\right]=\alpha$ ,

${v}_{o}\left[2\right]=\left(1-\alpha \right)\alpha +\alpha$ ,

${v}_{o}\left[3\right]=\left(1-\alpha {\right)}^{2}\alpha +\left(1-\alpha \right)\alpha +\alpha$ ,

${v}_{o}\left[k\right]=\alpha \sum _{k=0}^{n-1}\left(1-\alpha {\right)}^{k}=1-\left(1-\alpha {\right)}^{n}$

[We have made use of an important formula for the summation of a geometric series which we will prove shortly.] Hence, the solution is

${v}_{o}\left[n\right]=\left(1-\left(1-\alpha {\right)}^{n}\right)u\left[n\right]$

The solution

${v}_{o}\left[n\right]=\left(1-\left(1-\alpha {\right)}^{n}\right)u\left[n\right]$

is shown plotted below for $\alpha$ = 0.25.

4/ Important side issue — the summation of finite geometric series

Suppose

$S=\sum _{n=l}^{k}{a}^{n}={a}^{l}+{a}^{l+1}+\text{.}\text{.}\text{.}+{a}^{k}$

Then

$\mathrm{\alpha S}=\alpha \sum _{n=l}^{k}{a}^{n}={a}^{l+1}+{a}^{l+2}+\text{.}\text{.}\text{.}+{a}^{k+1}$

Hence,

$\left(1-\alpha \right)S=\left(1-\alpha \right)\sum _{n=l}^{k}{a}^{n}={a}^{l}-{a}^{k+1}$

and

$S=\sum _{n=l}^{k}{a}^{n}=\frac{{a}^{l}-{a}^{k+1}}{\left(1-\alpha \right)}$

Conclusion

• Difference equations arise in a variety of interesting contexts.
• Simple difference equations can be solved iteratively.
• We will develop more systematic and efficient methods to solve arbitrary difference equations.

X. GENERAL LINEAR DIFFERENCE EQUATION

For a DT system, such as the ones shown previously, we can write the relation between an input variable x[n] and an output variable y[n]by a general difference equation of the form

$\sum _{k=0}^{K}{a}_{k}y\left[n+k\right]=\sum _{l=0}^{L}{b}_{l}x\left[n+1\right]$

We seek a solution of this system for an input that is of the form

$x\left[n\right]={\text{Xz}}^{n}u\left[n\right]$

where X and z are in general complex quantities and u[n] is the unit step function. We will also assume that the system is at rest for n<0, so that y[n] = 0 for n<0.

We will seek the solution for n>0 by finding the homogeneous (unforced solution) and then a particular solution (forced solution).

1/ Homogeneous solution

a/ Geometric (exponential) solution

Let the homogeneous solution be ${y}_{h}\left[n\right]$ , then the homogeneous equation is

$\sum _{k=0}^{K}{a}_{k}y\left[n+k\right]=0$

To solve this equation we assume a solution of the form

${y}_{h}\left[n\right]={\mathrm{A\lambda }}^{n}$

Since

${y}_{h}\left[n+k\right]={\mathrm{A\lambda }}^{n+k}$

we have

$\sum _{k=0}^{K}{a}_{k}{\mathrm{A\lambda }}^{n+k}=0$

b/ Characteristic polynomial

The equation can be factored to yield

$\left(\sum _{k=0}^{K}{a}_{k}{\lambda }^{k}\right){\mathrm{A\lambda }}^{n}=0$

We are not interested in the trivial solution

${y}_{h}\left[n\right]={\mathrm{A\lambda }}^{n}=0$

Therefore, we can divide by ${\mathrm{A\lambda }}^{n}$ to obtain the characteristic polynomial

$\sum _{k=0}^{K}{a}_{k}{\lambda }^{k}=0$

This polynomial of order K has K roots which can be exposed by writing the polynomial in factored form

$\prod _{k=1}^{K}\left(\lambda -{\lambda }_{k}\right)=0$

c/ Natural frequencies

The roots of the characteristic polynomial $\left\{{\lambda }_{1},{\lambda }_{2},\text{.}\text{.}\text{.}{\lambda }_{K}\right\}$ are called natural frequencies. These are frequencies for which there is an output in the absence of an input.

If the natural frequencies are distinct, i.e., if ${\lambda }_{i}\ne {\lambda }_{k}$ for $i\ne k$ the most general homogeneous solution has the form

${y}_{h}\left[n\right]=\sum _{k=1}^{K}{A}_{k}{\lambda }_{{k}^{n}}$

Two-minute miniquiz problem

Problem 10-1 — Natural frequencies of the ladder network

The electric ladder network shown below has the following difference equation

$-{v}_{o}\left[n+1\right]+\left(2+\text{rg}\right){v}_{o}\left[n\right]-{v}_{o}\left[n-1\right]={\text{rgv}}_{i}\left[n\right]$

Find the natural frequencies.

Solution

The natural frequencies are determined by the homogeneous solution

$-{v}_{o}\left[n+1\right]+\left(2+\text{rg}\right){v}_{o}\left[n\right]-{v}_{o}\left[n-1\right]=0$

Substituting a solution of the form ${v}_{0}\left[n\right]={\mathrm{A\lambda }}^{n}$ results in

 $-{\mathrm{A\lambda }}^{n+1}+\left(2+\text{rg}\right){\mathrm{A\lambda }}^{n}-{\mathrm{A\lambda }}^{n-1}-1=0$

which yields the characteristic polynomial

${\lambda }^{2}-\left(2+\text{rg}\right)\lambda +1=0$

whose roots are the characteristic frequencies

${\lambda }_{1,2}=\left(1+\frac{\text{rg}}{2}\right)±\sqrt{\left(1+\frac{\text{rg}}{2}{\right)}^{2}-1}$

d/ Natural frequencies of the ladder network

The electric ladder network shown below has the following difference equation

$-{v}_{o}\left[n+1\right]+\left(2+\text{rg}\right){v}_{o}\left[n\right]-{v}_{o}\left[n-1\right]={\text{rgv}}_{i}\left[n\right]$

The natural frequencies are:

${\lambda }_{1,2}=\left(1+\frac{\text{rg}}{2}\right)±\sqrt{\left(1+\frac{\text{rg}}{2}{\right)}^{2}-1}$

The natural frequencies depend on rg,

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